Time reference circuits are often used in the arts where periodic waveforms or pulses are required. Many factors can affect the accuracy of such circuits such as inherent variations in resistance, capacitance, and other properties of the circuit components themselves. Variations in electrical properties to due changes in temperature can further complicate matters. For example, relaxation oscillators are one type of time reference circuits commonly used in monolithic IC designs. The basic mechanism is to charge or discharge a capacitor either through an R-C network or with a constant current source, thus producing a periodic waveform output. The output frequency is inversely proportional to the sum of the charging and the discharging times, and the duty cycle depends on the ratio of the charging to discharging time. Relaxation oscillators can be generally classified as R-C relaxation oscillators or constant-current relaxation oscillators. Simplified conceptual circuit schematic diagrams of examples of R-C and constant-current relaxation oscillators and their outputs are shown in FIG. 1A (prior art) through FIG. 2B.
Referring primarily to FIGS. 1A (prior art), and 1B, an example of an R-C relaxation oscillator circuit 10 is shown. The thresholds of the Schmitt trigger 12 are Va and Vb. The hysteresis of the Schmitt trigger 12 is Va-Vb. With the switch S1 open, vo1, rises exponentially toward Vcc. When it reaches Vb, it trips the Schmitt 12 and S1 closes, vo1 then drops exponentially toward VL. VL is as given by Equation 2. When vo1 reaches Va the Schmitt trigger 12 changes state and switch S1 is opened once again, repeating the previous cycle. The frequency of the R-C relaxation oscillator circuit 10 output VOUT1 shown in FIG. 1B can also be described by Equations 1 thru 7.
                              tau          ⁢                                          ⁢          1                =                  r          ⁢                                          ⁢          1          ⁢          c          ⁢                                          ⁢          1                                    (                  eq          ⁢                                          ⁢          1                )                                VL        =                              Vcc            .            r                    ⁢                                          ⁢                      2            /                          (                                                r                  ⁢                                                                          ⁢                  1                                +                                  r                  ⁢                                                                          ⁢                  2                                            )                                                          (                  eq          ⁢                                          ⁢          2                )                                          tau          ⁢                                          ⁢          2                =                                            (                              r                ⁢                                                                  ⁢                1                ⁢                                                                  ⁢                r                ⁢                                                                  ⁢                2                            )                        .            c                    ⁢                                          ⁢          1                                    (                  eq          ⁢                                          ⁢          3                )                                          t          ⁢                                          ⁢          1                =                  tau          ⁢                                          ⁢          1.          ⁢                      ln            ⁡                          (                              Vcc                -                                  Va                  /                  Vcc                                -                Vb                            )                                                          (                  eq          ⁢                                          ⁢          4                )                                          t          ⁢                                          ⁢          2                =                  tau          ⁢                                          ⁢          2.          ⁢                      ln            ⁡                          (                              Vb                -                                  VL                  /                  Va                                -                VL                            )                                                          (                  eq          ⁢                                          ⁢          5                )                                f        =                              1            /            T                    =                      1            /                          (                                                t                  ⁢                                                                          ⁢                  1                                +                                  t                  ⁢                                                                          ⁢                  2                                            )                                                          (                  eq          ⁢                                          ⁢          6                )                                          1          /          f                =                  r          ⁢                                          ⁢          1          ⁢          c          ⁢                                          ⁢          1.          ⁢                      {                                          ln                ⁡                                  (                                      Vcc                    -                                          Va                      /                      Vcc                                        -                    Vb                                    )                                            +                                                (                                      r                    ⁢                                                                                  ⁢                                          2                      /                                              (                                                                              r                            ⁢                                                                                                                  ⁢                            1                                                    +                                                      r                            ⁢                                                                                                                  ⁢                            2                                                                          )                                                                              )                                .                                  ln                  ⁡                                      (                                          Vb                      -                                              VL                        /                        Va                                            -                      VL                                        )                                                                        }                                              (                  eq          ⁢                                          ⁢          7                )            
Now referring primarily to FIGS. 2A (prior art) and 2B, an example of a constant-current relaxation oscillator circuit 20 is shown. It may be seen that the constant-current source oscillator 20 uses a current I2 instead of a resistor (R2 in FIG. 1A) to charge and discharge a timing capacitor C2. With I2 cut off, I1 charges C2 towards Vcc until vo2 equals Va. Then the Schmitt trigger 22 output goes high and the current source I2 is turned on. Assuming I2>I1, I2−I1 now discharges the capacitor until vo2 equals Vb. The Schmitt trigger 22 then changes state again and turns off I2 to repeat the cycle. FIG. 2B portrays an example of a waveform output from a constant-current oscillator such as in the circuit shown in FIG. 2A (prior art).
Commonly in monolithic designs, the only constant source available is a voltage VBG generated from a band-gap reference. The current IBG generated from a band-gap referenced voltage VBG using an on-chip resistor R makes the current IBG dependent on the variation of the resistor R. Equations 8 thru 13 describe the output frequency dependency of a constant-current source oscillator on the on-chip resistor R and capacitor C.
                              i          ⁢                                          ⁢          1                =                              Vbg            /            r                    ⁢                                          ⁢          1                                    (                  eq          ⁢                                          ⁢          8                )                                          i          ⁢                                          ⁢          2                =                              k            .            i                    ⁢                                          ⁢          1          ⁢                      (                          k              >              1                        )                                              (                  eq          ⁢                                          ⁢          9                )                                          t          ⁢                                          ⁢          1                =                              (                          Vb              -              Va                        )                    ⁢          c          ⁢                                          ⁢                      1            /            i                    ⁢                                          ⁢          1                                    (                  eq          ⁢                                          ⁢          10                )                                          t          ⁢                                          ⁢          2                =                                            (                              Vb                -                Va                            )                        .            c                    ⁢                                          ⁢                      1            /                          (                              k                -                1                            )                                ⁢          i          ⁢                                          ⁢          1                                    (                  eq          ⁢                                          ⁢          11                )                                f        =                              1            /            T                    =                                    1              /                              (                                                      t                    ⁢                                                                                  ⁢                    1                                    +                                      t                    ⁢                                                                                  ⁢                    2                                                  )                                      =                          i              ⁢                                                          ⁢              1.              ⁢                                                (                                      k                    -                    1                                    )                                /                                  (                                                            k                      .                      c                                        ⁢                                                                                  ⁢                    1.                    ⁢                                          (                                              Vb                        -                        Va                                            )                                                        )                                                                                        (                  eq          ⁢                                          ⁢          12                )                                f        =                              Vbg            .                          (                              k                -                1                            )                                /                      (                                          k                .                r                            ⁢                                                          ⁢              1.              ⁢              c              ⁢                                                          ⁢              1.              ⁢                              (                                  Vb                  -                  Va                                )                                      )                                              (                  eq          ⁢                                          ⁢          13                )            
The frequency of a relaxation oscillator depends on the R-C product and also on the supply voltage if the threshold of the trigger is dependent on the supply voltage. Circuit techniques have been developed to make the trip points or Schmitt trigger thresholds supply-independent. One way known in the arts to make the trip points supply-independent is to use band-gap voltage as a reference for comparison to the voltage across the capacitor.
An example of the band-gap referenced approach is illustrated in FIG. 3 (prior art). The relaxation oscillator circuit 30 of FIG. 3 uses a band-gap referenced current iBG. The current iBG is switched using two inverters INV1, INV2 to charge two capacitors C3, C4 alternately. The first capacitor C3 is charged to VBG, after which it trips the first comparator COMP1. The current iBG is then switched to charge the second capacitor C4. The output of the oscillator VOUT3 is high while the first capacitor C3 is charged, and it then goes low when the first comparator COMP1 trips. When the voltage V4 on the second capacitor C4 reaches VBG, a second comparator COMP2 trips, and the current iBG is switched back to the first capacitor C3, and so on. Thus, a periodic waveform output VOUT3 is obtained with a 50% duty cycle. The output frequency depends on how much time the current iBG takes to charge the capacitors C3, C4 to VBG. Note that VBG is independent of power supply. It is known to use external trim to correct the output frequency when for the effects of variation in temperature and process. Table 1 gives an example of representative frequency variation of this circuit 30 with two trim bits used to trim the frequency.
TABLE 1Trim = 01Trim = 01Trim = 11SupplyNOMINALWEAK fre-STRONGVTEMPFrequency (MHz)quency (MHz)Frequency (MHz)1.65−405.894.665.811.65275.394.205.361.651505.123.985.191.80−405.784.525.721.80275.344.165.321.801505.113.975.171.95−405.704.455.661.95275.304.135.291.951505.103.965.16
As shown in Table 1, the frequency is independent of supply voltage VCC because the charging current VBG and the reference voltage VBG do not change with supply voltage VCC. There is however, significant variation over temperature and process. Trim must be used to re-center the frequency over process, in the absence of which the frequency is as high as 7.6 MHz at the strong corner with supply being 1.65 V at cold temperatures. These variations are mainly due to variations in capacitor and resistor values. The resistor variation leads to variation in the constant current VBG that charges the capacitors, e.g. C3, C4, and causes frequency variation at the output. The value of the capacitors may change with process, which also causes frequency variation. In this example, in the absence of trim the oscillator output frequency varies from 7.6 to 3.96 MHz while it is centered around 5.5 MHz, a variation of 3.6 MHz.
The frequency of relaxation oscillators used in monolithic designs depends on the resistors and capacitors used in their implementation. The output frequency therefore varies to the extent that the resistors and the capacitors vary over process and temperature, and due to changes in the supply voltage. Though the supply dependency may be overcome using techniques known in the arts, the variation over process persists unless external trim components are used. Due to these and other problems, methods and circuits for providing relaxation oscillators insensitive to variations in process, supply voltage, or temperature changes would be useful and advantageous in the arts.